Question

Pipes \[A\] and $B$ can fill a tank in $5$ hours and $6$ hours respectively. Another pipe $C$ can empty the tank in $12$ hours. If all the three pipes are opened together, how much time will it take to fill the tank?

Answer 1

Hint:
The individual time needed for each pipe for filling or emptying is given. So we can calculate the part of the tank filled in one hour. Its reciprocal gives the time needed to fill the tank.

Complete step by step solution:
It is given that pipes \[A\] and $B$ can fill a tank in $5$ hours and $6$ hours respectively and another pipe $C$ can empty the tank in $12$ hours.
We are asked to find the time needed to fill the tank if the three tanks are opened together.
Since the pipe $A$ needs $5$ hours to fill the tank, we can say that $\dfrac{1}{5}$ of the tank is filled in one hour by $A$.
Since the pipe $B$ needs $6$ hours to fill the tank, we can say that $\dfrac{1}{6}$ of the tank is filled in one hour by $B$.
So in one hour, part of the tank filled by $A$ and $B$ is $\dfrac{1}{5} + \dfrac{1}{6}$.
Also since the pipe $C$ needs $12$ hours to empty the tank, we can say that $\dfrac{1}{{12}}$ of the tank is emptied in one hour by $C$.
Considering all these we can say that the part of tank filled in one hour when three pipes are open is $\dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{{12}}$.
This gives,
Tank filled in one hour $ = \dfrac{{12}}{{60}} + \dfrac{{10}}{{60}} - \dfrac{5}{{60}} = \dfrac{{12 + 10 - 5}}{{60}}$
Simplifying we get,
Tank filled in one hour $ = \dfrac{{17}}{{60}}$
Now we want to find the time needed to fill the tank.
Since the number of hours times the part of tank filled in one hour gives one, we have,
$t \times \dfrac{{17}}{{60}} = 1$, where $t$ is the number of hours.
So we get,
$t = \dfrac{{60}}{{17}}$
This can be written as $\dfrac{{60}}{{17}} = \dfrac{{51}}{{17}} + \dfrac{9}{{17}}$ which is equal to $3\dfrac{9}{{17}}$.
So the time needed to fill the tank if $A,B$ and $C$ are open is $3\dfrac{9}{{17}}$ hours.

$\therefore $ The answer is $3\dfrac{9}{{17}}$ hours.

Note:
This problem can also be done in an efficient method. Efficiency is the work done in unit time (here one hour).
Let us consider the capacity of the tank as $60$(LCM of $5,6,12$). The efficiency of the pipes can be considered as $12,10$ and $ - 5$ respectively. So, total efficiency is $12 + 10 - 5 = 17$.
We know Work done is the product of efficiency and time.
So time is gained by dividing the capacity (work done) by the total efficiency.
This gives, $time = \dfrac{{60}}{{17}}hours$